I have always liked Schroeder's take on the partition function being a product of translational and internal degrees of freedom:

$$ Z_1 = Z_{\text {trans}} Z_{\text{int}} $$ where $Z_{\text{int}}$ can be rotational, vibrational, electronic, etc.

Consider a homonulear (both end of the molecule are two of the same species of atom, i.e. indistinguishable) diatomic molecule confined to 2 dimensions BUT for the moment do not consider its rotation yet.

Is $Z_{\text{int}}=2$? This seems to fit with Schroeder's statement below the above definition on page 252 "An oxygen molecule, for example, has a threefold-degenerate ground state, which contributes a factor of 3 to the internal partition function." I was always confident in this $Z_{\text{int}}=2$ for situations like this, but once I DO consider rotation I am doubting my intuition.

Consider a pencil, eraser end and point end. If you spin it on your desk, there are X different orientations. But if the pencil had TWO points (like kids do in primary school) then you wouldn't be able to distinguish between previous distinct orientation, so there would be X/2 different orientations. This fits with the "$\sigma$ =2 for homonuclear molecules" from this site https://en.wikibooks.org/wiki/Statistical_Thermodynamics_and_Rate_Theories/Molecular_partition_functions#Rotational_Partition_Function

My confusion lies in why WITHOUT rotation, the partition function is multiplied by 2 to account for its indistinguishably, but WITH rotation we divide it by 2. Can someone help me grasp this?


Could be something like this?

Partition function for indistinguishable classical particles : $Z=\frac{1}{N!}\int dE\ g(E)e^{-\beta E}=\frac{1}{N!}\int d^Nq \ d^Np\ e^{-\beta H( \{ p,q \})}$

No rotation, two particles, 2D: $Z=\frac{1}{2!}\int d^2q \ d^2p \ e^{-\beta H_T( \{ p,q \})}= \frac{Z_T}{2}$

Two particles, 2D: $Z=\frac{1}{2!}\int d^2q \ d^2p \ e^{-\beta (H_T( \{ p,q \}+H_R( \{ p,q \}))}=\frac{1}{2} \int d^2q \ d^2p \ e^{-\beta (H_T( \{ p,q \})}e^{-\beta H_R( \{ p,q \})}=\frac{Z_T}{2} \int d^2q \ d^2p \ e^{-\beta (H_R( \{ p,q \})}=\frac{Z_T}{2} (\int dq \ dp \ e^{-\beta (H_R(p,q)})^2$

That gives the desired result if the partition function for the rotational hamiltonian of one particle equals two, $\int dq \ dp \ e^{-\beta H_R(p,q)}=2$


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